Solve {l}{4y=2x-8}{5x-4y=20}

$$\left. \begin{array} { l } { 4 y = 2 x - 8 } \\ { 5 x - 4 y = 20 } \end{array} \right.$$

Solve for y, x

$$x=4$$

$$y=0$$

Steps Using Substitution

Consider the first equation. Subtract 2x from both sides.

$$4y-2x=-8$$

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$4y-2x=-8,-4y+5x=20$$

Choose one of the equations and solve it for y by isolating y on the left hand side of the equal sign.

$$4y-2x=-8$$

Add 2x to both sides of the equation.

$$4y=2x-8$$

Divide both sides by 4.

$$y=\frac{1}{4}\left(2x-8\right)$$

Multiply \frac{1}{4} times -8+2x.

$$y=\frac{1}{2}x-2$$

Substitute \frac{x}{2}-2 for y in the other equation, -4y+5x=20.

$$-4\left(\frac{1}{2}x-2\right)+5x=20$$

Multiply -4 times \frac{x}{2}-2.

$$-2x+8+5x=20$$

Add -2x to 5x.

$$3x+8=20$$

Subtract 8 from both sides of the equation.

$$3x=12$$

Divide both sides by 3.

$$x=4$$

Substitute 4 for x in y=\frac{1}{2}x-2. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=\frac{1}{2}\times 4-2$$

Multiply \frac{1}{2} times 4.

$$y=2-2$$

Add -2 to 2.

$$y=0$$

The system is now solved.

$$y=0,x=4$$

Steps Using Matrices

Consider the first equation. Subtract 2x from both sides.

$$4y-2x=-8$$

Put the equations in standard form and then use matrices to solve the system of equations.

$$4y-2x=-8,-4y+5x=20$$

Write the equations in matrix form.

$$\left(\begin{matrix}4&-2\\-4&5\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}-8\\20\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&-2\\-4&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}4&-2\\-4&5\end{matrix}\right))\left(\begin{matrix}4&-2\\-4&5\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}4&-2\\-4&5\end{matrix}\right))\left(\begin{matrix}-8\\20\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}4&-2\\-4&5\end{matrix}\right))\left(\begin{matrix}-8\\20\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}4&-2\\-4&5\end{matrix}\right))\left(\begin{matrix}-8\\20\end{matrix}\right)$$

For the 2\times 2 matrix \left(\begin{matrix}a&b\\c&d\end{matrix}\right), the inverse matrix is \left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right), so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{5}{4\times 5-\left(-2\left(-4\right)\right)}&-\frac{-2}{4\times 5-\left(-2\left(-4\right)\right)}\\-\frac{-4}{4\times 5-\left(-2\left(-4\right)\right)}&\frac{4}{4\times 5-\left(-2\left(-4\right)\right)}\end{matrix}\right)\left(\begin{matrix}-8\\20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{5}{12}&\frac{1}{6}\\\frac{1}{3}&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}-8\\20\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{5}{12}\left(-8\right)+\frac{1}{6}\times 20\\\frac{1}{3}\left(-8\right)+\frac{1}{3}\times 20\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}0\\4\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=0,x=4$$

Steps Using Elimination

Consider the first equation. Subtract 2x from both sides.

$$4y-2x=-8$$

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$4y-2x=-8,-4y+5x=20$$

To make 4y and -4y equal, multiply all terms on each side of the first equation by -4 and all terms on each side of the second by 4.

$$-4\times 4y-4\left(-2\right)x=-4\left(-8\right),4\left(-4\right)y+4\times 5x=4\times 20$$

Simplify.

$$-16y+8x=32,-16y+20x=80$$

Subtract -16y+20x=80 from -16y+8x=32 by subtracting like terms on each side of the equal sign.

$$-16y+16y+8x-20x=32-80$$

Add -16y to 16y. Terms -16y and 16y cancel out, leaving an equation with only one variable that can be solved.

$$8x-20x=32-80$$

Add 8x to -20x.

$$-12x=32-80$$

Add 32 to -80.

$$-12x=-48$$

Divide both sides by -12.

$$x=4$$

Substitute 4 for x in -4y+5x=20. Because the resulting equation contains only one variable, you can solve for y directly.

$$-4y+5\times 4=20$$

Multiply 5 times 4.

$$-4y+20=20$$

Subtract 20 from both sides of the equation.

$$-4y=0$$

Divide both sides by -4.

$$y=0$$

The system is now solved.

$$y=0,x=4$$